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Consider the partial differential equation

(rho + nu) V[y, x] == 
  Log[1 / (1 + y (x - y) (beta(psi - D[V[y, x], y]) - eps D[V[y, x], x]))] - 
  eps y (x - y) D[V[y, x], x]

with the boundary condition

V[y, y] == 0

rho, nu, beta, psi, eps are strictly positive, known constants.

I want to solve for V[y, x] over the domain y <= x, 0 <= y <= 1, 0.6 <= x <= 1.

We pose the following problem to Mathematica:

par = 
  {rho -> 0.00014, nu -> 0.00182, psi -> 227.1, beta -> 0.0966, eps -> 0.01;

set = 
  {(rho + nu) V[y, x] == 
     Log[1/(1 + y (x - y) (beta (psi - D[V[y, x], y]) - eps D[V[y, x], x]))] - 
     eps y (x - y) D[V[y, x], x] , V[y, y] == 0} /. par;

Reg = ImplicitRegion[(y <= x), {{y, 0, 1}, {x, 0.6, 1}}];

NDSolve[set, V, {y, x} ∈ Reg]

I got the error message:

NDSolve::conarg: The arguments should be ordered consistently.

I don't understand this message.

bbgodfrey explains that Mathematica might not be able to solve problems with such a boundary condition. Following bbgodfrey's suggested approach I therefore try to solve the equation without the boundary condition, using NDSolveValue (because bbgodfrey uses DSolveValue). We pose the problem

NDSolveValue[
  (rho + nu) V[y, x] == 
    Log[1/(1 + y (x - y) (beta (psi - D[V[y, x], y]) - eps D[V[y, x], x]))] - 
    eps y (x - y) D[V[y, x], x] /. par, V, {y, x} ∈ Reg]

This yields the error message

NDSolveValue::femper: PDE parsing error of {0.00196701 V + 0.01 V$3966 (x - y) y - Log[1/(1 + (Times[<<2>>] + Times[<<2>>]) (x + Times[<<2>>]) y)]}. Inconsistent equation dimensions.

I am a beginner as far as PDEs in Mathematica (and StackExchange) are concerned. Please give suggestions on how to proceed.

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  • $\begingroup$ The second formulation, without the boundary conditions, generates some warnings but it does produce a result in my MMA (v. 12.0 on WIn10-64). I do not get the femper parsing error. $\endgroup$
    – MarcoB
    Commented Jan 14, 2021 at 20:33
  • 2
    $\begingroup$ You can specify your boundary condition like this: DirichletCondition[V[y, x] == 0, x == y]. But the solution doesn't converge. $\endgroup$
    – Michael E2
    Commented Jan 14, 2021 at 22:21
  • $\begingroup$ This equation is not quasilinear therefore accordingly to general PDE theory it should be turn to the quasilinear form first. $\endgroup$
    – Alex Trounev
    Commented Jan 15, 2021 at 15:09

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